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I've been reading this article https://en.wikipedia.org/wiki/Celestial_coordinate_system and a question came to my mind.

The minimum units are seconds ( ″ ). I picture an angle of 1 second as being like a "cone" going out from Earth. It's really small at the beginning but as you go further its width increases.

How wide does this cone get as you get further into the universe? How big is an arcsecond on different astronomical bodies?

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  • $\begingroup$ The coordinate system is as precise as you want to make it. It's like any other coordinate system: add more decimal places and you have a more precise location. $\endgroup$ – Asher Apr 11 '17 at 15:47
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    $\begingroup$ What @Asher says is not exactly true. The precision of your coordinate system is limited by how precisely you can define its zero-point. However the positions of things in this coordinate system can have finite precision, which is determined by your telescope/measurment method. $\endgroup$ – AtmosphericPrisonEscape Apr 11 '17 at 22:26
  • $\begingroup$ so it depends on the telescope you are using? how can that be the rule? $\endgroup$ – Dan Wears Prada Apr 12 '17 at 7:19
  • $\begingroup$ Edited to reflect the comment below my answer. $\endgroup$ – James K Apr 13 '17 at 8:28
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How big is one arcsecond at various distances?

An arcsecond is a small angle, 1/3600 of a degree or about 5 millionths of a radian ($4.85\times10^{-6}$). To estimate the size of something that appears 1 arcsecond across you can use the small angle approximation to trigonometry: Multiply the distance to the object by $4.85\times10^{-6}$

Examples:

  • One arcsecond on the moon is 1.87km
  • One arcsecond on the sun is 727km
  • One arcsecond on Mars (when it is closest to Earth) is 237km
  • One arcsecond at distance of one parsec is one astronomical unit (AU), by definition
  • One arcsecond on Alpha Centauri is 200 million km
  • One arcsecond on the Andromeda galaxy is 100 trillion km

The atmosphere limits how much detail you can see. Typically the smallest detail you can see is about 3 or 4 arcseconds, though professional telescopes do better by being built on the top of mountains and using various tricks, like adaptive optics.

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  • $\begingroup$ Ys this is exactly what i was looking for thanks! Now i feel we are really short-sighted... I will edit my question to your bold $\endgroup$ – Dan Wears Prada Apr 13 '17 at 8:06
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James K's answer is right on, and probably what you're after. I just wanted to mention an effect which comes into play if you look really, really far away:

Because light moves at a finite speed, we see galaxies (and other things) as they were in the past. And because the Universe expands, everything was closer together in the past, so we see distant galaxies as they were when they were closer and hence spanned a larger angle on the sky.

Thus, we have two competing effects: On the one hand, galaxies become smaller and smaller with distance, as expected and as decribed in James K's answer. On the other hand, we look further and further into the past, and hence see galaxies at a time when they looked larger and larger. The result — which is obtained by integrating the Friedmann equation — is that galaxies become smaller and smaller out to a certain distance (roughly 15 billion lightyears), after which the second effect starts to dominate and they start to grow in size.

Or, in other words, an arcseconds spans a larger and larger physical size (as described by James K) out to ~15 Gly (where it spans roughly 28,000 lightyears), after which it spans a smaller and smaller size. The relation looks like this:

scale

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  • $\begingroup$ I understand that we see the galaxies as they were in the past, so we see them closer than their actual distance to us. But i dont unserstand the last paragraph :/ $\endgroup$ – Dan Wears Prada Apr 13 '17 at 8:44
  • $\begingroup$ @DanWearsPrada: I edited the last two paragraphs. Do they make sense now? $\endgroup$ – pela Apr 13 '17 at 20:02
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    $\begingroup$ That is very interesting. Thank you $\endgroup$ – YaDa Jul 22 at 17:43
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You can do your own calculations easily, no need to ask anyone. Just change the distance units in the examples below:

tan(1 arcsec) * 10 km

tan(1 arcsec) * 4 light years

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  • $\begingroup$ To be 100% exact, shouldn’t it be 2*tan(0.5 arcsec)*distance? But for all practical purposes, approximating it by the arc as in James K’s answer is probably sufficient. $\endgroup$ – chirlu Apr 14 '17 at 9:10

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